cp's OEIS Frontend

As a flat sequence, this is a permutation of the nonnegative integers with inverse A368240.

See A353293 for the other k's.

Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).

In order to construct this sequence as a triangle we use the following rules:

- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.

- Then replace this new set of parts by the concatenation of its parts.

- Then replace this string by its mirror version which is a binary number.

T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).

The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.

First differs from A065609 at a(23).

Conjecture: this sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]

Row sums of triangle A128937. - Philippe Deléham, May 02 2007

a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009

a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011

For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012

a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015

From Gus Wiseman, Oct 30 2022: (Start)

Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).

Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.

(End)

Characterization: Let e_{p}(m) denote the exponent of the prime p in the prime factorization of m and [.] denote the Iverson bracket, then

e_{p}(a(n)) = Sum_{k>=1} [floor(n/p^k) is even].

This implies, among other things, that no prime > floor(n/2) can divide a(n). The prime exponents e_{2}(a(2n)) give Guy Steele's sequence GS(5,3) A080100.

Asymptotics: log a(n) ~ n(1 - log 2). It is conjectured that log a(n) ~ n(1 - log 2) + O(n^{1/2+eps}) for all eps > 0.

Bounds: A056040(floor(n/3)) <= a(n) <= A056040(floor(n/2)) if n >= 285.

Guy Steele defines a family of 36 integer sequences, denoted here by GS(i,j) for 1 <= i, j <= 6, as follows. a[1]=1; a[2n] = i-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}; a[2n+1] = j-th term of {0,1,a[n],a[n]+1,2a[n],2a[n]+1}. The present sequence is GS(1,5).

The full list of 36 sequences:

GS(1,1) = A000007

GS(1,2) = A000035

GS(1,3) = A036987

GS(1,4) = A007814

GS(1,5) = A135416 (the present sequence)

GS(1,6) = A135481

GS(2,1) = A135528

GS(2,2) = A000012

GS(2,3) = A000012

GS(2,4) = A091090

GS(2,5) = A135517

GS(2,6) = A135521

GS(3,1) = A036987

GS(3,2) = A000012

GS(3,3) = A000012

GS(3,4) = A000120

GS(3,5) = A048896

GS(3,6) = A038573

GS(4,1) = A135523

GS(4,2) = A001511

GS(4,3) = A008687

GS(4,4) = A070939

GS(4,5) = A135529

GS(4,6) = A135533

GS(5,1) = A048298

GS(5,2) = A006519

GS(5,3) = A080100

GS(5,4) = A087808

GS(5,5) = A053644

GS(5,6) = A000027

GS(6,1) = A135534

GS(6,2) = A038712

GS(6,3) = A135540

GS(6,4) = A135542

GS(6,5) = A054429

GS(6,6) = A003817

(with a(0)=1): Moebius transform of A038712.

From Trevor Green (green(AT)snoopy.usask.ca), Nov 26 2003: (Start)

a(n)/n has an accumulation point at x exactly when x is in the interval [1, 2]. Proof: Clearly n <= a(n) < 2n. Let b(n) = a(n)/n, then b(n) must always lie in [1,2) and all the accumulation points of the sequence must lie in [1,2]. We shall show that every such number is an accumulation point.

First, consider any d-bit integer n. Suppose that z of these bits are 0. Let n' be the (d+z)-bit integer whose first d bits are the same as those of n and whose remaining bits are all 1. Then a(n') will have to be the (d+z)-bit integer whose first d bits are all 1 and whose last z bits are all 0.

Thus n' = (n+1)*2^z-1; a(n') = (2^d-1)2^z; and b(n') = (2^d-1)/(n+1) + epsilon, where 0 < epsilon < 2^(1-d). So to get an accumulation point x, we just choose n(d) to be the d-bit integer such that (2^d-1)/(n(d)+1) < x <= (2^d-1)/n(d), or equivalently, n(d) = floor((2^d-1)/x). If x lies in [1,2), then n(d) will always be a d-bit number for sufficiently large d.

Then n'(d) yields an increasing subsequence of the integers for which b(n'(d)) converges to x. For x = 2, choose n(d) = 2^(d-1), which is always a d-bit number; then b(n'(d)) = (2^d-1)/(2^(d-1)+1) + epsilon = 2 + epsilon', where epsilon' also heads for 0 as d blows up. This proves the claim.

(End)

A080100(n) = number of numbers k such that n AND k = 0 in n-th row of the triangular array.